Question: $\begin{aligned} &f(x)=9x^2 \\\\ &g(x)=\dfrac{\sqrt{12-x}}{3} \end{aligned}$ $(f\circ g) (-4)=$
Explanation: Let's start by rewriting $(f\circ g) (-4)$ as $f(g(-4))$. When evaluating composite functions, we work our way inside out. To evaluate $f(g(-4))$, let's first evaluate $g(-4)$. Then we'll plug that result into $f$ to find our answer. Let's evaluate $g({-4})$. $\begin{aligned}g(x)&=\dfrac{\sqrt{12-x}}{3}\\\\ g({-4})&=\dfrac{\sqrt{12-({-4})}}{3}~~~~~~~~~~\text{Plug in }x={-4}\\\\ &=\dfrac{\sqrt{16}}{3}\\\\ &={\dfrac{4}{3}}\end{aligned}$ We now know that $f(g({-4}))$ is the same as $f\left({\dfrac{4}{3}}\right)$ because $g({-4}) = {\dfrac{4}{3}}$. Let's evaluate $f\left({\dfrac{4}{3}}\right)$. $\begin{aligned}f(x)&=9x^2\\\\ f\left({{\dfrac{4}{3}}}\right)&=9\left({\dfrac{4}{3}}\right)^2~~~~~~~~~~\text{Plug in }x={\dfrac{4}{3}}\\\\ &=9\cdot \left(\dfrac{16}{9}\right)\\\\ &=16\end{aligned}$ The answer: $(f\circ g)(-4) =16$